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Chapter 4: Problem 21

What volume of \(0.2500 \mathrm{M}\) cobalt(III) sulfate is required to react completely with (a) \(25.00 \mathrm{~mL}\) of \(0.0315 \mathrm{M}\) calcium hydroxide? (b) \(5.00 \mathrm{~g}\) of sodium carbonate? (c) \(12.50 \mathrm{~mL}\) of \(0.1249 M\) potassium phosphate?

Short Answer

Expert verified
Answer: The volumes of a 0.2500 M cobalt(III) sulfate solution needed to react with the given substances are: (a) 2.1 mL, (b) 94.352 mL, and (c) 12.49 mL.

Step by step solution

01

(1) Determine the chemical formula of cobalt(III) sulfate and given compounds

Cobalt(III) sulfate has the chemical formula \(\mathrm{Co_2(SO_4)_3}\). The given compounds have the chemical formulas: calcium hydroxide (Ca(OH)2), sodium carbonate (Na2CO3), and potassium phosphate (K3PO4).
02

(2) Write the balanced chemical equations for the reactions with cobalt(III) sulfate

(a) \(2\mathrm{Co_2(SO_4)_3 + 3Ca(OH)_2 \rightarrow 6CaSO_4 + 6H_2O}\) (b) \(\mathrm{Co_2(SO_4)_3 + 2Na_2CO_3 \rightarrow 4Na_2SO_4 + Co_2(CO_3)_3}\) (c) \(2\mathrm{Co_2(SO_4)_3 + K_3PO_4 \rightarrow 6K_2SO_4 + 2Co_3(PO_4)_2}\)
03

(3) Determine the stoichiometry ratios for the reactions

From the balanced equations, we have the following stoichiometry ratios: (a) 2 moles of \(\mathrm{Co_2(SO_4)_3}\) reacts with 3 moles of \(\mathrm{Ca(OH)_2}\). (b) 1 mole of \(\mathrm{Co_2(SO_4)_3}\) reacts with 2 moles of \(\mathrm{Na_2CO_3}\). (c) 2 moles of \(\mathrm{Co_2(SO_4)_3}\) reacts with 1 mole of \(\mathrm{K_3PO_4}\).
04

(4) Calculate the moles required for each reactant and use stoichiometry ratios to find the moles of cobalt(III) sulfate

(a) Moles of \(\mathrm{Ca(OH)_2}\) (molarity * volume in liters) = \(0.0315 \mathrm{M} \times (25.00 \mathrm{mL} / 1000) = 0.0007875 \mathrm{mol}\). From the stoichiometry: 2 moles of \(\mathrm{Co_2(SO_4)_3}\) reacts with 3 moles of \(\mathrm{Ca(OH)_2}\), so moles of \(\mathrm{Co_2(SO_4)_3}\) required = \((2/3) \times 0.0007875 \mathrm{mol} = 0.000525 \mathrm{mol}\). (b) Moles of \(\mathrm{Na_2CO_3}\) (mass / molar mass) = \(\frac{5.00 \mathrm{g}}{105.988 \mathrm{g/mol}} = 0.047176 \mathrm{mol}\). From the stoichiometry: 1 mole of \(\mathrm{Co_2(SO_4)_3}\) reacts with 2 moles of \(\mathrm{Na_2CO_3}\), so moles of \(\mathrm{Co_2(SO_4)_3}\) required = \((1/2) \times 0.047176 \mathrm{mol} = 0.023588 \mathrm{mol}\). (c) Moles of \(\mathrm{K_3PO_4}\) (molarity * volume in liters) = \(0.1249 \mathrm{M} \times (12.50 \mathrm{mL} / 1000) = 0.00156125 \mathrm{mol}\). From the stoichiometry: 2 moles of \(\mathrm{Co_2(SO_4)_3}\) reacts with 1 mole of \(\mathrm{K_3PO_4}\), so moles of \(\mathrm{Co_2(SO_4)_3}\) required = \((2/1) \times 0.00156125 = 0.0031225 \mathrm{mol}\).
05

(5) Calculate the volume of each cobalt(III) sulfate solution

(a) Volume of \(\mathrm{Co_2(SO_4)_3}\) solution: \(\frac{0.000525 \mathrm{mol}}{0.2500 \mathrm{M}} = 0.0021 \mathrm{L}\) or \(2.1 \mathrm{mL}\). (b) Volume of \(\mathrm{Co_2(SO_4)_3}\) solution: \(\frac{0.023588 \mathrm{mol}}{0.2500 \mathrm{M}} =0.094352 \mathrm{L}\) or \(94.352 \mathrm{mL}\). (c) Volume of \(\mathrm{Co_2(SO_4)_3}\) solution: \(\frac{0.0031225 \mathrm{mol}}{0.2500 \mathrm{M}} =0.01249 \mathrm{L}\) or \(12.49 \mathrm{mL}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Reaction Equations
Understanding how chemical reaction equations work is like learning the language of chemistry. They tell us exactly what and how much of each substance reacts to form new products. Here's a basic run-through:

In every balanced chemical equation, the number of atoms for each element must be equal on both sides of the reaction. This conservation of mass is crucial because atoms are neither created nor destroyed during a chemical reaction. The balanced equations from the exercise (e.g., 2Co2(SO4)3 + 3Ca(OH)2 → 6CaSO4 + 6H2O) ensure the ratio of reactants to products is accurate, meaning for every two moles of cobalt(III) sulfate, three moles of calcium hydroxide will react completely. For students, getting comfortable with balancing equations is a must - it'll be the foundation for all your stoichiometry calculations!
Molarity and Concentration
Diving into molarity and concentration, we're talking about how 'strong' or 'concentrated' a solution is. Imagine you're making lemonade - the amount of lemon juice you add determines how strong it tastes. Similarly, in chemistry, molarity (M) measures the amount of solute (like cobalt(III) sulfate in our example) divided by the volume of the solution (mostly water), usually in liters.

So, when we see something like 0.2500 M, it means there are 0.2500 moles of the substance in one liter of solution. Calculating the volume of reactants, like in the exercise, hinges on understanding molarity. If you see a solution labeled with a molarity, just remember, it's giving you a clue about how much stuff is packed into each liter of liquid, and that's going to be essential in working out your reaction stoichiometry.
Mole-to-Mole Ratios
And now to the pivot of stoichiometry — the mole-to-mole ratios. These ratios stem from the balanced chemical reaction equations and tell us how reactants pair up with each other. Just like a recipe that tells you how many eggs to mix with flour for pancakes, mole ratios inform us how many moles of one chemical react with another.

For instance, in the exercise when we see that 2 moles of Co2(SO4)3 react with 3 moles of Ca(OH)2, we have a 2:3 ratio. This ratio is the essential bridge between reactants in a recipe. Without these, it would be near impossible to determine the exact amounts needed for a reaction to take place completely. Get these ratios right, and you're well on your way to mastering stoichiometry!

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Most popular questions from this chapter

What is the volume of \(0.885 M\) hydrochloric acid required to react with (a) \(25.00 \mathrm{~mL}\) of \(0.288 \mathrm{M}\) aqueous ammonia? (b) \(10.00 \mathrm{~g}\) of sodium hydroxide? (c) \(25.0 \mathrm{~mL}\) of a solution \(\left(d=0.928 \mathrm{~g} / \mathrm{cm}^{3}\right)\) containing \(10.0 \%\) by mass of methylamine \(\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right) ?\)

Decide whether a precipitate will form when the following solutions are mixed. If a precipitate forms, write a net ionic equation for the reaction. (a) potassium nitrate and magnesium sulfate (b) silver nitrate and potassium carbonate (c) ammonium carbonate and cobalt(III) chloride (d) sodium phosphate and barium hydroxide (e) barium nitrate and potassium hydroxide

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For an acid-base reaction, what is the reacting species (the ion or molecule that appears in the chemical equation) in the following bases? (a) barium hydroxide (b) trimethylamine \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}\) (c) aniline, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) (d) sodium hydroxide

Assign oxidation numbers to each element in (a) methane (b) sulfurous acid (c) sodium oxide (d) the dihydrogen phosphate ion \(\left(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\right)\)
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